Solve the equation. $\dfrac{dy}{dx}=2x\cdot e^{-y}$ Choose 1 answer: Choose 1 answer: (Choice A) A $y=\ln(x^2+C)$ (Choice B) B $y=\ln(x^2)+C$ (Choice C) C $y=-\ln(-x^2+C)$ (Choice D) D $y=-\ln(-x^2)+C$
Explanation: We can bring this equation to the form $f(y)\,dy=g(x)\,dx$ : $\begin{aligned} \dfrac{dy}{dx}&=2x\cdot e^{-y} \\\\ \dfrac{dy}{dx}&=\dfrac{2x}{e^y} \\\\ e^y\,dy&=2x\,dx \end{aligned}$ This means we can solve this equation using separation of variables! $\begin{aligned} e^y\,dy&=2x\,dx \\\\ \int e^y\,dy&=\int 2x\,dx \\\\ e^y&=x^2+C \\\\ \ln(e^y)&=\ln(x^2+C) \\\\ y&=\ln(x^2+C) \end{aligned}$ Notice that after the integration, more work was required in order to isolate $y$. In conclusion, this is the solution of the equation: $y=\ln(x^2+C)$